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3 years ago

PLEASE HELP FAST! my problem is in the photo.

Mathematics

1 answer:

Xelga [282]3 years ago

7 0

[c] 70°

40+2x=180
2x=140
x=70

i use 2x since the angle measures of D and C are equivalent since it is an isosceles

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What is 27+3÷ 3 +4-1+72.4-34.24+6-4+75÷5 <br> Answer this fast please!

Dimas [21]

Answer:

86.16

Step-by-step explanation:

27+3÷ 3 +4-1+72.4-34.24+6-4+75÷5

PEMDAS

Multiply and divide first

3/3 is 1 and 75/5 = 15

Replace these into the equation

27+1 +4-1+72.4-34.24+6-4+15

Then add and subtract from left to right

31 +72.4-34.24+6-4+15

69.16+6-4+15

86.16

7 0

3 years ago

Read 2 more answers

Find several points that are on the graph of the function y = x^2 + 1. Plot the points in a coordinate plane.

beks73 [17]

Answer:

Below.

Step-by-step explanation:

Plot following points.

Calculate the point by plugging in values of x into x^2 + 1

for example When x = 0, y = 0^1 + 1 = 1.

So plot plot (0, 1),

Make a table of points to plot:

x -3 -2 -1 0 1 2 3

y 10 5 2 1 2 5 10

When you plot the points you'll see the graph is U shaped.

The function is of second degree (as it contains x^2) so it wont be linear.

8 0

3 years ago

Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randoml

Temka [501]

<h2>Answer:</h2>

The probability that both televisions work is: 0.42

The probability at least one of the two televisions does not work is:

0.5833

<h2>Step-by-step explanation:</h2>

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

9-3=6

The probability that both televisions work is calculated by:

$=\dfrac{6_C_2}{9_C_2}$

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

$=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42$

The probability at least one of the two televisions does not work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

$=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5$